Conic Sections Class 11
Master Conic Sections Class 11 with comprehensive NCERT Solutions, Practice Questions, MCQs, Sample Papers, Case Based Questions, and Video lessons.
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Conic Sections Class 11 – NCERT Solutions
Each question below opens its complete step-by-step Teachoo solution.
Ex 10.1
15 questionsEx 10.1, 1
Ex 10.1, 1 teachoo.com
Find the equation of the circle with centre (0, 2) and radius 2
Equation of a circle is
(x-h)}* + (y -—k)? = r? (1)
Where (h, k) is the centre & r is the radius
Here
Centre (h, k) = (0, 2}
So, h=0,k=2
& radius (r} = 2
Putting the values in (1)
(x- 0)? + (y - 2)? = 2?
Ex 10.1, 2
Ex 10.1, 2 teachoo.com
Find the equation of the circle with centre (-2, 3) and radius 4
We know that equation of a circle is
(x-h)+(y-kP =r? wa)
Where (h, k) is the centre & r is the radius
Here
Centre (h, k) = (—2, 3)
Soh=-2&k=3
& radius (r) = 4.
Putting values in (1)
(x - (-2))? + (y- 3)? = (4?
(x+ 2)? + (y-3)/ =16
(x2 + 22 + 4x) + (y? + 3? - Gy) = 16
Ex 10.1, 3
Ex 10.1, 3 teachoo.com
+ + : + 11 : 1
Find the equation of the circle with centre Gz) and radius D
We know that equation of a circle is
(x-h)?+(y-k)?=r? — ...(1)
Where (h, k) is the centre & r is the radius
Here
1
Centre (h, k) = G. ;)
1
Soh=-&k==
2 4
+ 1
& Radius =r=—.
12
Ex 10.1, 4
Ex 10.1, 4 teachoo.com
Find the equation of the circle with centre (1, 1) and radius v2
We know that equation of a circle is
(x—h)? + (y —k}? =r? (1)
Where (h, k) is the centre & r is the radius
Here
Centre (h, k) = (1, 1)
Soh=1&k=1
& radius (r) = v2.
Putting values in (1)
(x- 1) + (y— 1) = (v2)?
Ex 10.1, 5
Ex 10.1, 5 teachoo.com
Find the equation of the circle with centre (—a, -b) & radius Va? — b?
We know that equation of a circle is
(x-h)?+(y-kP =r? (1)
Where (h, k) is the centre & r is the radius
Here
Centre (h, k) = (-a, -b)
Soh=-a&k=-b
& Radius r= Va? —b?*.
Putting values in (1)
(x= (-a))? + (y - (-b))? = (Va? — BY
Ex 10.1, 6
Ex 10.1, 6 teachoo.com
Find the centre and radius of the circle (x + 5)? + (y— 3)? = 36
Given (x +5)? + (y—3)? = 36
(x -(-5))? + (y- 3)? = 36 (1)
We know that
Circle of equation is given by
(x-—h)? +(y-k}? =r?
Comparing (1) & (2)
h=-5,k=3, r°=36
Thus, centre of circle = (h, k)
=(-5, 3)
& Radius =r=6
Ex 10.1, 7
Ex 10.1, 7 teachoo.com
Find the centre and radius of the circle x? + y? - 4x - 8y -45 =0
Given x? + y?- 4x - 8y -45 =0. .(1)
We need to make this in form
(x—h)? + (y—k)? =r? (2)
From (1)
X°- 4x+y’- By =45
(x? — 2 (x) (2)) + (y?- 2 (y) (4)) = 45
[x2 — 2(x)(2) + 22-27] + [y? - 2(y)(4)+ 4?- 47] = 45
[x? — 2(x)(2) + 2] + [y= 2(y)(4}+ 4] = 2?— 4? = 45
Using (a - b}? = a? + b? - 2ab
Ex 10.1, 8
Ex 10.1, 8 teachoo.com
Find the centre and radius of the circle x? + y* -— 8x + 10y-12=0
Given x? + y*— 8x + 10y - 12 =0. (1)
We need to make this in form
(x-h)? +(y—k) =r? (2)
From (1)
x? +y*- 8x + 10y-12 =0
x? -8xt+y?+10y-12=0
(x? — 8x) + (y?+ 10y) - 12 =0
[x? — 2(x)(4)] + [y? + 2(y)(5)] - 12 = 0
[x2 — 2(x)(4) + 42 - 47] + [y?+ 2(y)(5) + 52 - 57) -12=0
Ex 10.1, 9
Ex 10.1, 9 teachoo.com
Find the centre and radius of the circle 2x? + 2y?-x =0
Given 2x? + 2y*-x=0 (1)
We need to make this in form
(x-—h)? + (y-k)? =r?
From (1)
2x? + 2y*-x=0
2ay2— ~\e
2(x +y’ *\ =0
24y_~ = 0
+p sa5
22 x 2=
x 2 +y°=0
Ex 10.1, 10
Ex 10.1, 10 teachoo.com
Find the equation of the circle passing through the points (4, 1) and
(6, 5) and whose centre is on the line 4x + y = 16.
Let the equation of the circle be
(x-hP +(y-—kP =r’. w (A)
Since circle passes through (4, 1) | Since circle passes through (6, 5)
It will satisfy the equation of circle | It will satisfy the equation of circle
Putting x = 4 & y = Lin (A) Putting x = 6, & y=Sin (A)
_ 2. _ 2 2
(xh)? 4 (y—k)2=r2, (x-hP +(y-kPe=r
6—h)?+ (5—k)*=r?
(4-h)?+(1-k)2=r? ( e+e
6? +h? - 12h + 52 +k? - 10k =r?
4 +h?-8h+12+k? -2k=r?
36+ h?- 12h+ 25 +k?- 10k =r?
2 22 =p
16 +h’ Bh+1 +k? 2k=r h2+k?- 12h- 10k +36 +25=P
W+k?—Bh-2k +1750 (1) | p24 k2-12h-10k+61=r2 ... (2)
Ex 10.1, 11
Ex 10.1, 11 teachoo.com
Find the equation of the circle passing through the points (2, 3) and
(-1, 1) and whose centre is on the line x — 3y —11 =0.
Let the equation of the circle be
(x-h)? + (y-k)? =r? (A)
Since the circle passes through points (2, 3)
Point (2, 3) will satisfy the equation of circle
Putting x = 2, y = 3 in (A)
(2—h)? + (3—k)? =r?
(2)? + (hy? - 2(2)(h) + (3)? + k? - 2(3)(k) = 7
4+h?-4h+9+k?-6k=r’
h?+k?- 4h-6k+4+9=97
h? +k?- 4h - 6k + 13 =r? .. (1)
Ex 10.1, 12
Ex 10.1, 12 (Method 1) teachoocom
Find the equation of the circle with radius 5 whose centre lies on x-
axis and passes through the point (2, 3).
We know that equation of circle is
(2, 3)
(xh) + (y—kP =?
Centre of circle is denoted by (h, k)
Since it lies on x-axis ,
k=0
Hence
Centre of circle = (h, 0)
& given Radius =5
Ex 10.1, 13
Ex 10.1, 13 teachoo.com
Find the equation of the circle passing through (0, 0) and making
intercepts a and b on the coordinate axes.
Let the equation of circle be
(RP + (y-kP=P (0)
where (h, k) is the centre b \
& ris the radius of a circle ol, “Ala, 0)
Also given that circle making intercepts
a & bon the coordinate axes
Let intercept on x-axis be a,
So, coordinates of point A (a, 0)
Ex 10.1, 14
Ex 10.1, 14 (Method 1) teachoo.com
Find the equation of a circle with centre (2, 2) and passes through
the point (4, 5).
We know that equation of circle is
(x—h)?4 (y-kp=r? (4,5)
Since, centre of circle is (2, 2)
Soh=2&k=2
Our equation becomes
(x- 2? +(y-2P =r? (1)
Now,
Distance between centre and point on circle = radius
Distance between points (h, 0) & (2,3) =r
V¥(4-2)?+(5 -—2)%=r
Ex 10.1, 15
. teachoo.com
Ex 10.1, 15 (Introduction)
Does the point (—2.5, 3.5) lie inside, outside or on the circle x? + y? = 2
We know that equation of circle is
(xh) + (y—kP =r?
(a, b)
where (h, k) is the centre |
& ris the radius of circle Le J
b
If for any point (a, b) (a, b)
1. (a-h)?+(b-k)2=r?
Then point (a, b) lies on the circle
2. (a-h)*+(b-k)? <r?
Then point (a, b) lie inside the circle
Ex 10.2
12 questionsEx 10.2, 1
Ex 10.2, 1 teachoo.com
Find the coordinates of the focus, axis of the parabola, the
equation of directrix and the length of the latus rectum for y? = 12x
Given y? = 12x
Directrix | Coefficient Latus
Equation Focus
Equation of x rectum
x-axis y= 4ax = (a, 0) xX=-a + 4a
x-axis y?=-4ax (-a, 0) x=a - 4a
y-axis x2=4ay (0, a) y=-a + 4a
y-axis x?=-4ay (0, -a) y=a - 4a
Since the above equation is involves y?
Its axis is x-axis
Ex 10.2, 2
Ex 10.2, 2 teachoo.com
Find the coordinates of the focus, axis of the parabola, the
equation of directrix and the length of the latus rectum for x? = 6y
Given equation is x? = 6y
Directrix | Coefficient Latus
Equation of x rectum
x-axis y?=4ax = (a, 0) X=-a + 4a
x-axis y?=-4ax (-a, 0) x=a - 4a
y-axis x?=4ay (0,a) y=-a + 4a
y-axis x?=-4day (0, -a) y=a - 4a
Since the above equation is involves x?
Its axis is y-axis
Ex 10.2, 3
Ex 10.2, 3 teachoo.com
Find the coordinates of the focus, axis of the parabola, the equation
of directrix and the length of the latus rectum for y? =—- 8x
Given equation is y? =—8x.
Axis Directrix | Coefficient Latus
Equation of x rectum
x-axis y?=4ax (a, 0) X=-a + 4a
x-axis y?=-4ax (-a, 0) X=a - da
y-axis x?=4ay (0,a) y=-a + 4a
y-axis x*=~4ay (0, -a) y=a - 4a
Since the above equation is involves y*
Its axis is x-axis
Ex 10.2, 4
Ex 10.2, 4 teachoo.com
Find the coordinates of the focus, axis of the parabola, the equation
of directrix and the length of the latus rectum for x? =-16y
Given equation is x? = -16y.
Directrix | Coefficient Latus
Equation of x rectum
x-axis y?=4ax (a,0) x=-a + 4a
xaxis y?=-dax (-a,0) x=a - 4a
y-axis x?=4ay (0,a) y=-a + 4a
y-axis x?=-4ay (0,-a) y=a - 4a
Since the above equation is involves x?
Its axis is y-axis
Also coefficient of y is negative (-)
Hence we use equation
x? = - day
Ex 10.2, 5
Ex 10.2, 5 teachoo.com
Find the coordinates of the focus, axis of the parabola, the equation
of directrix and the length of the latus rectum for y* = 10x
Given equation is y? = 10x.
Directrix | Coefficient
Equation of x
x-axis y? = 4ax (a,0) x=-a + 4a
x-axis y?=-4ax — (-a, 0) x=a - 4a
y-axis x? = 4ay (0,a) y=-a + 4a
y-axis x*=-day (0, -a) y=a - 4a
Since the above equation is involves y*
Its axis is x-axis
Ex 10.2, 6
Ex 10.2, 6 teachoo.com
Find the coordinates of the focus, axis of the parabola, the equation
of directrix and the length of the latus rectum for x? = —9y
Given equation is x? = —9y.
Directrix | Coefficien
Equation tof x
x-axis y?=4ax = (a, 0) x=-a + 4a
x-axis y?=-4ax (-a, 0) xX=a - 4a
y-axis x? = Jay (0, a) y=-a + 4a
y-axis x2=-Gay (0, -a) y=a - 4a
Since the above equation is involves x?
Its axis is y-axis
Ex 10.2, 7
Ex 10.2, 7 teachoo.com
Find the equation of the parabola that satisfies the following
conditions: Focus (6, 0); directrix x = -6
Since focus lies on x-axis
Hence equation is either y* = 4ax or y? = -4ax
Now focus has positive x co-ordinate
So, we have to use equation y? = 4ax
Coordinate of focus = (a, 0)
(a, 0) = (6, 0)
a=6
Hence equation of parabola is
y? = 4ax
Ex 10.2, 8
Ex 10.2, 8 teachoo.com
Find the equation of the parabola that satisfies the following
conditions: Focus (0, —3); directrix y = 3
Since focus lies on y-axis
Hence equation is either x* = 4ay & x? =—4ay
Now focus has a negative y co-ordinate
So, we have to use equation x? = -4ay
Coordinate of focus = (0, —a)
(0, -a) = (0, -3)
-a=-3
a=3
Ex 10.2, 9
Ex 10.2, 9 teachoo.com
Find the equation of the parabola that satisfies the following
conditions: Vertex (0, 0); focus (3, 0)
Since focus lies on y-axis
Hence equation is either y* = 4ax & y? = - 4ax
Now focus has a negative y co-ordinate
So, we have to use equation y? = 4ax
Coordinate of focus = (a, 0)
(a, 0) = (3, 0)
a=3
Equation of parabola is
y? = 4ax
Ex 10.2, 10
Ex 10.2, 10 teachoo.com
Find the equation of the parabola that satisfies the following
conditions: Vertex (0, 0) focus (-2, 0)
Since focus lies on y-axis
Hence equation is either y* = 4ax & y? = -4ax
Now focus has a negative y co-ordinate
So, we have to use equation y” = -4ax
Coordinate of focus = (-a, 0)
(-a, 0) = (-2, 0)
-a=-2
a=2
Ex 10.2, 11
Ex 10.2, 11 teachoo.com
Find the equation of the parabola that satisfies the following
conditions: Vertex (0, 0) passing through (2, 3) and axis is along x-axis
Given that axis is along the x-axis
So, equation of parabola is of the form
y’ = 4ax or y? = -4ax
(2, 3)
Plotting point (2, 3)
Since point (2, 3) lie in the 1 quadrant
& parabola passes through the point (2, 3)
The parabola will be of the form +
Ex 10.2, 12
Ex 10.2, 12 teachoo.com
Find the equation of the parabola that satisfies the following
conditions: Vertex (0, 0) passing through (5, 2) and axis is along y-axis
Given that axis is along the y-axis,
So, equation of parabola is of the form
x =4ay orx?=-day
Plotting point (5, 2) \ (5, 2)
Since point lie of the 1* quadrant
& parabola passes through the point (5, 2)
So, the parabola is of the form \ y
Ex 10.3
20 questionsEx 10.3, 1
Ex 10.3, 1 teachoo.com
Find the coordinates of the foci, the vertices, the length of major
axis, the minor axis, the eccentricity and the length of the latus
; ey
rectum of the ellipse stiet 1
The given equation is
2 y2 (1)
are eee |
36 16
Coordinat | Vertices Eccent | Latus
es of foci ricity§ | Rectum
x oy 4 xaxis C=a*-b? (tc,0) (4a,0) 2a 2b e=£ 2b?
az b2 a a
ana 1y-axis C?=a?-b? (Otc) (0,4 a) 2a 2b e=* 2b?
b2 a a a
Since 36 > 16,
The above equation is of the form
xe y? 2
atl (2)
Ex 10.3, 2
Ex 10.3, 2 teachoo.com
Find the coordinates of the foci, the vertices, the length of major
axis, the minor axis, the eccentricity and the length of the latus
: arn aa
rectum of the ellipse att 1
Coy
at2s. 1 (1)
Coordinat | Vertices Eccent
es of foci ricity
o+Ge1 xaxis c?=a2-b? (tc,0) (+a, 0) 2a 2b e=< 20?
a
o+ts 1 y-axis c?=a?-b? (0c) (0,4. a) 2a 2b e=< 20?
a
Since 4< 25
The above equation is of the form
x? y?
pe + 2 =1 .(2)
Ex 10.3, 3
Ex 10.3, 3 teachoo.com
Find the coordinates of the foci, the vertices, the length of major
axis, the minor axis, the eccentricity and the length of the latus
; ey
rectum of the ellipse tot 1
2 2
74% 21 (1)
16° 9
Coordinat | Vertices Eccent
es of foci ricity
o+Ge1 xaxis c?=a?-b? (£¢,0) = (+a, 0) 2a 2b e=s 20?
a
yd yayig C=ar—-b? (0,4 _¢ 2p?
witial yaxis C=a b2 (0,4 c) (0,+ a) 2a 2b en
Since 16 >9
Hence the above equation is of the form
x? y?
ata (2)
Ex 10.3, 4
Ex 10.3, 4 teachoo.com
Find the coordinates of the foci, the vertices, the length of major
axis, the minor axis, the eccentricity and the length of the latus
; ey
rectum of the ellipse a3 i007 1
x? y?
33007 1 (1)
Coordinat | Vertices Eccent | Latus
es of foci ricity§ | Rectum
o+Ge1 x-axis c?=a2-b? (tc,0) (+a, 0) 2a 2b e=< 20?
a
x? y? . _ a2 € 2b2
sataz=l yaxis c@=at—b? (0,t¢) (0,+ a) 2a 2b e= c=
Since 25 < 100
Hence the above equation is of the form
x? y?
ptgot (2)
Ex 10.3, 5
Ex 10.3, 5 teachoo.com
Find the coordinates of the foci, the vertices, the length of major
axis, the minor axis, the eccentricity and the length of the latus
ey
rectum of the ellipse — + —=1
49 36
x? y? _
2936, 1 (1)
in
. Vertices os,
of foci tricity | Rectum
voy x-axis c?=a2-b2 (4c,0) (4a,0) 2a 2b e=<* 2b?
a2 p2 a a
wv y-axis c2=a2—b? (Otc) (0,ta) 2a 2b e=* 2b?
be a2 a a
Since 49 > 36
Hence the above equation is of the form
x? y? _
tpl ...{2)
Ex 10.3, 6
Ex 10.3, 6 teachoo.com
Find the coordinates of the foci, the vertices, the length of major
axis, the minor axis, the eccentricity and the length of the latus
; ey
rectum of the ellipse toot a007 1
x? y?
100 * 400 ~ 1 (1)
Coordinat | Vertices Eccent | Latus
es of foci ricity§ | Rectum
a ae x-axis c2=a2-b? (tc,0) (+a, 0) 2a 2b exh 2H
@ b2 a a
x? y? . _ a2 € 2b2
satig=l yaxis c=a2-b? (0,4c) (0,+ a) 2a 2b e= = =
Since 100 < 400
Hence the above equation is of the form
x? y?
7 + ae 1 .(2)
Ex 10.3, 7
Ex 10.3, 7 teachoo.com
Find the coordinates of the foci, the vertices, the length of major
axis, the minor axis, the eccentricity and the length of the latus
rectum of the ellipse 36x? + 4y? = 144
Given
36x? + Ay? = 144.
Dividing equation by 144
2 2
36x" By 1)
144 144
Coordinat | Vertices Eccent
es of foci ricity
o+Ge1 x-axis c?=a?—-b? (+c,0) (+a, 0) 2a 2b e=* 20?
a a
cn ae 1 y-axis c?=a?-b? (0,4) (0,+ a) 2a 2b e=£ 2b?
D2 a a a
Since 4 < 36
Above equation is of form
x? y?
wats (2)
Ex 10.3, 8
Ex 10.3, 8 teachoo.com
Find the coordinates of the foci, the vertices, the length of major
axis, the minor axis, the eccentricity and the length of the latus
rectum of the ellipse 16x? + y? = 16
Given
16x? + y* = 16.
Dividing equation by 16
16 2 2
16x" y" _ 16
16°16 16
2
247 e
x + 16 1
x? y?
ptpttl w(L)
Coordinat | Vertices Eccent | Latus
es of foci ricity | Rectum
E+5e1 x-axis c?=a2-b2 (tc,0) (+a, 0) 2a 2b e=< 2b?
a
44 yaxis @=at-b? (O40) (0,4a) 2a 2b e=£ 2”
b2 a2 a a
Ex 10.3, 9
Ex 10.3, 9 teachoo.com
Find the coordinates of the foci, the vertices, the length of major axis,
the minor axis, the eccentricity and the length of the latus rectum of
the ellipse 4x? + 9y* = 36
Given
4x? + Sy? = 36.
Divide equation by 36
2 2
ax, oy" _ 36 (1)
36 3636
Coordinat | Vertices Eccent
es of foci ricity
44 axis @=at-b? (+60) (+a, 0) 2a 2b e=f
@ pb a a
aoe ae) y-axis c?=a2-b? (0,4) (0,+ a) 2a 2b e=f 22
D2 a a a
Since 9 > 4
Hence the above equation is of the form
x? y?
3 + ae 1 .(2)
Ex 10.3, 10
teachoo.com
Ex 10.3, 10
Find the equation for the ellipse that satisfies the given conditions:
Vertices (+5, 0), foci (+4, 0)
Given
Vertices (+5, 0) (1)
Since the vertices are of form (+a, 0) (2)
Hence, Major axis is along x-axis
and equation of ellipse is
x? y?
a3 + po 1
From (1) & (2)
a=5
Ex 10.3, 11
Ex 10.3, 11 teachoo.com
Find the equation for the ellipse that satisfies the given conditions:
Vertices (0, +13), foci (0, +5)
Given Vertices (0, +13) {1}
Hence
The vertices are of the form (0, +a) (2)
Hence, the major axis is along y-axis
& Equation of ellipse is of the form
2 y2
= + Y 1
b2 a2
Ex 10.3, 12
Ex 10.3, 12 teachoo.com
Find the equation for the ellipse that satisfies the given conditions:
Vertices (+6, 0), foci (+4, 0)
Given
Vertices (+ 6, 0) (1)
The vertices are of the form (ta, 0) ...(2)
Hence, the major axis is along x-axis
& Equation of ellipse is of the form
x2 y2
w + rr =1
Ex 10.3, 13
Ex 10.3, 13 teachoo.com
Find the equation for the ellipse that satisfies the given conditions:
Ends of major axis (+3, 0), ends of minor axis (0, +2)
We need to find equation of ellipse
Given that
End of major axis = (+ 3, 0)
We know that
Ends of major axis are the vertices of the ellipse.
Vertices of the ellipse = (+3, 0) (1)
Vertices of the ellipse is of the form (+ a, 0) (2)
So required equation of ellipse is
x? y?
a3 + pe 1
Ex 10.3, 14
Ex 10.3, 14 teachoo.com
Find the equation for the ellipse that satisfies the given conditions:
Ends of major axis (0, + V5) , ends of minor axis (+1, 0)
Given ends of Major Axis (0, + V5),
& ends of Minor Axis (+1, 0)
Major axis is along the y-axis
So, our required equation of ellipse is
2 2
cao ae |
b2 a2
We know that
End of major axis is the vertices of the ellipse
Ex 10.3, 15
Ex 10.3, 15 teachoo.com
Find the equation for the ellipse that satisfies the given conditions:
Length of major axis 26, foci (+5, 0)
Given foci = (+ 5, 0) (1)
Since the foci is of the type (+c,0) —_...(2)
So the major axis is along the x-axis
& required equation of ellipse is
x? y?
a3 + po 1
From (1) & (2)
c=5
Ex 10.3, 16
Ex 10.3, 16 teachoo.com
Find the equation for the ellipse that satisfies the given conditions:
Length of minor axis 16, foci (0, +6)
We need to find equation of ellipse
whose length of minor axis = 16
& Foci = (0, +6) ...(1)
Since foci is of the type (0, +c) (2)
The major axis is along the y-axis.
& required Equation of Ellipse is
x2 2
pe + a =1
Ex 10.3, 17
Ex 10.3, 17 teachoo.com
Find the equation for the ellipse that satisfies the given
conditions: Foci (+3, 0),a=4
Given Foci (+3, 0) (1)
The foci are of the form (+c, 0) (2)
Hence the major axis is along x-axis
& equation of ellipse is of the form
xe y?
zz + po 1
From (1) on (2)
c=3
Also, givena=4
Ex 10.3, 18
Ex 10.3, 18 teachoo.com
Find the equation for the ellipse that satisfies the given conditions:
b = 3, c =4, centre at the origin; foci on the x axis.
We need to find equation of ellipse
Given b = 3, c = 4, centre at the origin
& foci on the x axis.
Since foci are on the x-axis
So, foci are of the form (+ c, 0}
And major axis is along x-axis &
Required equation of ellipse is
2 2
= +51
az pb?
Ex 10.3, 19
Ex 10.3, 19 teachoo.com
Find the equation for the ellipse that satisfies the given
conditions: Centre at (0, 0), major axis on the y-axis and passes
through the points (3, 2) and (1, 6).
Since major axis is along y-axis & centre is at (0,0)
So required equation of ellipse is
x2 2
pet ge At)
Given that ellipse passes through point (3, 2) & (1, 6)
Points (3, 2) & (1, 6) will satisfy equation of ellipse.
Ex 10.3, 20
Ex 10.3, 20 teachoo.com
Find the equation for the ellipse that satisfies the given conditions:
Major axis on the x-axis and passes through the points (4, 3) and (6, 2)
Since Major axis is on the x-axis
So required equation of ellipse is
x2 2
Given that ellipse passes through point (4, 3) & (6, 2)
Points (4, 3) & (6, 2) will satisfy the equation of ellipse
Ex 10.4
15 questionsEx 10.4, 1
teachoo.co
Ex 10.4, 1 AEROOLOM
Find the coordinates of the foci and the vertices, the eccentricity,
> yp
and the length of the latus rectum of the hyperbola ~ - 7 =1
Given equation is
Pra y*
Equationof Axis of Coordinate | Coordinate | Eccentri | Latus
Hyperbola Hyperbola of foci of vertice city rectum
2 y2 2
=i 1s xaxis C=a+b2 (+c, 0) (ta, 0) e=* =
ye x? , =f 2b?
ptt ovaxis = c= a?+b? —(0, 4c) (0, +a) e-7 =
The above equation is of the form
xe y?
@ pt (2)
Thus, axis of hyperbola is x-axis
Ex 10.4, 2
Ex 10.4, 2 teachoo.com
Find the coordinates of the foci and the vertices, the eccentricity,
and the length of the latus rectum of the hyperbola r - x =1
Given equation is
vey (1)
9° 27
wens etc [oan saa
Hyperbola | Hyperbola of foci of vertice city rectum
=-u- 1s xaxis C=a2+b2 (tc, 0) (ta, 0) e=< au
v-Ss1 yaxis c?=a?+b? (0, tc) (0, +a) e=e a
The above equation of hyperbola is of the form
-. Axis of Hyperbola is y-axis
Ex 10.4, 3
Ex 10.4, 3 teachoo.com
Find the coordinates of the foci and the vertices, the eccentricity,
and the length of the latus rectum of the hyperbola 9y? — 4x? = 36
The given equation is
Sy? — 4x? = 36
Divide whole equation by 36
Oy? — 4x7 36
3636
oy? 4
36.36% (1)
wits
4.9
Equation of | Axis of Coordinate | Coordinate | Eccentri | Latus
Hyperbola | Hyperbola of foci of vertice city rectum
2 y2 2
=e 1 sxaxis C=a+b2 = (+c, 0) (ta, 0) e=< =
ye x? . =f 2b?
opti yaxis | c?=a?+b? (0, +c} (0, ta) ern =
Ex 10.4, 4
Ex 10.4, 4 teachoo.com
Find the coordinates of the foci and the vertices, the eccentricity, and
the length of the latus rectum of the hyperbola 16x? — 9y? = 576
The given equation is
16x? — Sy* = 576.
Dividing whole equation by 576
16x? 9y? 576
576 576 576
x2 y
nF
36. 6A ..(1)
Equation of | Axis of Coordinate | Coordinate | Eccentri | Latus
Hyperbola | Hyperbola of foci of vertice city rectum
2 2 2
= - 3 =1 xaxis c?=a2+b? (tc, 0) (+a, 0) e= £ =
Cc
yx 24 yaxis c?=a?+b? = (0, +c) (0, +a) e=c 2b?
a@ b2 a
Ex 10.4, 5
Ex 10.4, 5 teachoo.com
Find the coordinates of the foci and the vertices, the eccentricity,
and the length of the latus rectum of the hyperbola 5y? — 9x? = 36
Given equation is
5y? — 9x? = 36.
Dividing whole equation by 36
Sy? 9x? 36
36 36 «36
2 2
y x
rns |
®) A (1)
5
Equation of | Axis of Coordinate | Coordinate | Eccentri | Latus
Hyperbola | Hyperbola of foci of vertice city rectum
2 y2 2
=-ie 1 sxaxis c=a2+b2 (tc, 0) (+a, 0) e=< =
ye x? . =f 2b?
opti yaxis c?=a2+b? (0, +c} (0, ta) ern =
Ex 10.4, 6
Ex 10.4, 6 teachoo.com
Find the coordinates of the foci and the vertices, the eccentricity, and
the length of the latus rectum of the hyperbola 49y? — 16x? = 784
49y? — 16x? = 784
Dividing whole equation by 784
pe.
784 784 784
yxy
16 49
ye?
popu wa)
Equation of | Axis of Coordinate | Coordinate | Eccentri | Latus
Hyperbola | Hyperbola of foci of vertice city rectum
2 y2 2
=i 1 _syaxis c=a2+b2 ~— (+c, 0) (ta, 0) e=* =
ye x? , =f 2b?
ptt ovaxis = c= a?+b? —(0, 4c) (0, +a) e-7 =
So our equation is of the form
y? x? _
@ pes 1 (2)
Ex 10.4, 7
Ex 10.4, 7 teachoo.com
Find the equation of the hyperbola satisfying the given
conditions: Vertices (+2, 0), foci (+3, 0)
Given Vertices are (+2, 0)
Hence, vertices are on the x-axis
-. Equation of hyperbola is of the form
x2 y*
eet
Now, Co-ordinate of vertices = (+a, 0) & Vertices = (+2, 0)
«. (ta, O) = (+2, 0)
Hence
a=2
Ex 10.4, 8
Ex 10.4, 8 teachoo.com
Find the equation of the hyperbola satisfying the given conditions:
Vertices (0, +5), foci (0, +8)
We need to find equation of hyperbola
given Vertices (0, +5), foci (0, +8)
Since Vertices are on the y-axis
So required equation of hyperbola is
e-Ee1
a B
We know that
Vertices =(0, +a)
Given Vertices = (0, +5)
So a=5
Ex 10.4, 9
teachoo.com
Ex 10.4, 9
Find the equation of the hyperbola satisfying the given conditions:
Vertices (0, +3), foci (0, +5)
We need to find equation of hyperbola
Given Vertices (0, +3), foci (0, +5)
Since Vertices are on the y-axis
So required equation of hyperbola is
2 Z
-. Axis of hyperbola is y-axis
Ex 10.4, 10
Ex 10.4, 10 teachoo.com
Find the equation of the hyperbola satisfying the given conditions:
Foci (+5, 0), the transverse axis is of length 8.
Co-ordinates of foci is (+5, 0)
Which is of form (+c, 0)
Hencec=5
Also , foci lies on the x-axis
So, Equation of hyperbola is
x2 y
@ pt
Ex 10.4, 11
Ex 10.4, 11 teachoo.com
Find the equation of the hyperbola satisfying the given
conditions: Foci (0, +13), the conjugate axis is of length 24.
We need to find equation of hyperbola
Given foci (0, +13) & conjugate axis is of length 24.
Since foci is on the y-axis
So required equation of hyperbola is
y* x2
@ pt
Now, Co-ordinates of foci = (0, + c) & given foci = (0, +13)
So, (0, + c) = (0, +13)
Ex 10.4, 12
Ex 10.4, 12 teachoo.com
Find the equation of the hyperbola satisfying the given conditions:
Foci (+ 3¥5, 0), the latus rectum is of length 8.
Co-ordinates of Foci is (+35, 0)
Since foci is on the x-axis
Hence equation of hyperbola is of the form
x2 y
Pr
Also,
We know that co-ordinates of foci are (+c, 0)
So, (+3V5, 0) = (+e, 0)
Ex 10.4, 13
Ex 10.4, 13 teachoo.com
Find the equation of the hyperbola satisfying the given conditions:
Foci (+4, 0), the latus rectum is of length 12
Since the foci are on the x-axis.
Hence, the required equation of the hyperbola is
x2 y*
@ wt
Now, coordinates of foci are (+c, 0) & given foci = (+4, 0)
so, (+c,0) = (+4,0)
c=4
Ex 10.4, 14
Ex 10.4, 14 teachoo.com
Find the equation of the hyperbola satisfying the given conditions:
Vertices (+7, 0), e ==
Here, the vertices are on the x-axis.
Therefore, the equation of the hyperbola is of the form
x2 y*
@ pt
Now, coor#tdinates of vertices are (+ a,0) & Given vertices = (+7, 0),
So, (+ a,0) = (+7, 0),
a=7
Ex 10.4, 15
Ex 10.4, 15 teachoo.com
Find the equation of the hyperbola satisfying the given conditions:
Foci (0,+V 10), passing through (2, 3)
Since Foci is on the y-axis
So required equation of hyperbola is
y x2
@ pt
Now, Co-ordinates of foci = (0, + c) & given foci = (0, +V10)
So, (0, +c) = (0, +V¥10)
c=¥10
Examples
20 questionsExample 1
Example 1 teachoo.com
Find an equation of the circle with centre at (0,0) and radius r
We know that equation of a circle is
(x-h)? + (y-kP =P (1)
Where (h, k) is the centre & ris the radius
Here
Centre (h, k) = (0, 0)
Soh=0&k=0
& radius =r.
Putting values in (1)
(x— 0) + (y— OP = (rP
Example 2
Example 2 teachoo.com
Find the equation of the circle with centre (-3, 2) and radius 4.
We know that equation of a circle is
(x —h)* + (y —k)? = r? (1)
Where (h, k) is the centre & ris the radius
Here
Centre (h, k) = (-3, 2)
Soh=-3&k=2
& radius (r) = 4.
Putting values in (1)
(x - (-3))? + (y- 2)? = (4?
Example 3
Example 3 teachoo.com
Find the centre and the radius of the circle x + y2 + 8x + 10y-8 =0
Given equation of circle
x +y?+8x+ 10y-8=0 (1)
We need to make this in form
(x-hpP + (y-—kP? =P?
From (1)
x+y? + 8x+10y-8=0
(x2 + 8x) + (y? + 10y) = 8
(x)? + 2(4)(x) + y° + 2(5){y) = 8
[x? + 2(4){(x) + (4)? - (4)7] + [y* + 2(5){y) + (5)? - (5)7] = 8
Example 4
Example 4 teachoo.com
Find the equation of the circle which passes through the points
(2, -2), and (3,4) and whose centre lies on the line x + y = 2.
Equation of circle with centre (h, k) is
(x —h)* + (y —k)? = r? w (A)
Since the circle passes through (2, — 2)
Point (2, -2) will satisfy the equation of circle
Putting x = 2, y = -2 in (A)
(2-h)? + (-2-kP =r
4+h?- 4h +4 + (-2)? + (-k)? + 2(-2)(-k) = r?
4+h?-4h+4+k? +4k=r?
h? +k? -4h+4k+8=r? w (1)
Example 5
Example 5 teachoo.com
Find the coordinates of the focus, axis, the equation of the
directrix and latus rectum of the parabola y? = 8x.
Given equation is y? = 8x.
Se
Equation of x
x-axis y?=4ax (a, 0) x=-a + 4a
x-axis y*=4ax (-a, 0) x=a - 4a
y-axis x?=4ay (0,a) y=-a + 4a
y-axis x?=-4ay (0, -a) y=a - 4a
Since the above equation involves y”
Its axis is x-axis
Also coefficient of x is positive (+8)
Hence, we use equation
y? = 4ax
Example 6
Example 6 teachoo.com
Find the equation of the parabola with focus (2, 0) and directrix
x=-2,
Since focus lies on x-axis
Hence equation is either y* = 4ax or y? = -4ax
Now, focus has positive x co-ordinate
So, we have to use equation y? = 4ax
Coordinates of focus = (a, 0)
(2, 0) = (a, 0)
Example 7
Example 7 teachoo.com
Find the equation of the parabola with vertex at (0, 0) & focus at (0, 2).
Since focus (0, 2} lie on the y-axis
Hence, Equation of parabola is
either x? = day & x? = -day
Since focus has positive y — co-ordinate
So, we have to use equation x? = 4ay
Coordinate of focus = (0, a)
(0, a) = (0, 2)
a=2
Example 8
Example 8 teachoo.com
Find the equation of the parabola which is symmetric about the
y-axis, and passes through the point (2,-3).
Since the parabola is symmetric about y-axis
Equation is of the form
x? = Gay or x? =— day
Now plotting point (2, -3) on graph
(2,-3)
Since (2, -3) lies in fourth quadrant
So the parabola will be of the form 7Y
Example 9
teachoo.co
Example 9 m
Find the coordinates of the foci, the vertices, the length of major
axis, the minor axis, the eccentricity and the latus rectum of the
. x y?
X,Y.
ellipse sto 1
(1)
. x? y?
eyes
Given at 9 1
Coordinat Vertices Eccent Latus
es of foci ricity Rectum
ce ae x-axis c2=a2-b? (tc,0) (+a, 0) 2a 2b enh 7
@ b2 a a
cae ae) y-axis c?=a?-b? (0,tc) = (0,4 a) 2a 2b e=f 2
b2 a2 a a
Since 25 >9
Hence the above equation is of the form
xe y?
zz + po 1
Example 10
Example 10 teachoo.com
Find the coordinates of the foci, the vertices, the lengths of major
and minor axes and the eccentricity of the ellipse 9x? + 4y* = 36.
Given
9x? + Ay? = 36
Dividing whole equation by 36
9x? +4y? 36
36036
9 4y?
2g
36 36
2 2
4 9
Coordinat Vertices Eccent Latus
es of foci ricity Rectum
we cy axis 2=a?-b? (4c,0) (ta,0) 2a 2b en 2
az b2 a a
pci y-axis C=a*-b? (Otc) (0,ta) 2a 2% eat 2
b2 az a a
Example, 11
teachoo.com
Example 11
Find the equation of the ellipse whose vertices are (+ 13, 0) and
foci are (+ 5, 0)
Given vertices are (+ 13, 0} (1)
The given vertices are of the form (+a, 0) (2)
Hence the major axis is along x-axis &
Equation of ellipse is of the form
xe y?
zz + po 1
From (1) & (2)
a=13
Example 12
teachoo.com
Example 12
Find the equation of the ellipse, whose length of the major axis is
20 and foci are (0, + 5).
We need to find equation of ellipse
Whose length of major axis = 20
& foci are (0, + 5) (1)
Since the foci are of the type (0, +c) w(2}
So the major axis is along the y-axis
& required equation of ellipse is
x2 2
pe + a =1
Example 13
Example 13 teachoo.com
Find the equation of the ellipse, with major axis along the x-axis and
passing through the points (4, 3) and (— 1,4).
Given that
Major axis is along x-axis
So required equation of ellipse is
x2 2
1 + Be =1 (1)
Given that point (4, 3) & (-1, 4) lie of the ellipse
So, point (4, 3} & (-1, 4) will satisfy equation of ellipse
Example 14 (i)
Example 14 teachoo.com
Find the coordinates of the foci and the vertices, the eccentricity,
the length of the latus rectum of the hyperbolas:
yey
()>-erh
The given equation is
2 2.
9 16
Equation of | Axis of Coordinate | Coordinate | Eccentri | Latus
Hyperbola | Hyperbola of foci of vertice city rectum
x? y? . 2b?
az pant x axis c=a2+b2 (tc, 0) (ta, 0) e=£ —
ye , _ + =£ 2b?
agent ovanis C=a%tb? (0, tc) (0, ta) e=c =
The above equation is of the form
x2 y*
sS-SF (2
a pb 1 (2)
Example 14 (ii)
Example 14
Find the coordinates of the foci and the vertices, the eccentricity, the length of the latus rectum of the hyperbolas:
(ii) y2 – 16x2 = 16
Example 15
Example 15 teachoo.com
Find the equation of the hyperbola with foci (0, + 3) and vertices
(0,
2
Since, foci are on the y-axis
So required equation of hyperbola is
2 Z
We know that
Vertices = (0, +a)
Given vertices are (0 =)
So, (0, ta) = (0 =)
Example 16
Example 16 teackoo.com
Find the equation of the hyperbola where foci are (0, +12) and the
length of the latus rectum is 36.
We need to find equation of hyperbola
Given foci (0, +12) & length of latus rectum 36.
Since foci is on the y-axis
So required equation of hyperbola is
2 2
x’
x ~y2=1
a b
Now, Co-ordinates of foci = (0, + c) & given foci = (0, +12)
Example 17
Example 17 teachoo.com
The focus of a parabolic mirror as shown in Fig 11.33 is at a distance
of 5 cm from its vertex. If the mirror is 45 cm deep, find the distance
AB (Fig 11.33). Y
45cm
= A
Since axis of parabola is in positive x-axis,
Equation of parabola is
y? = 4ax
Here,
a = Distance of focus from vertex = 5 cm 5
Now,
Mirror is 45 cm Deep
Hence, x-coordinate of point A = 45
x=45
Example 18
Example 18 teachoo.com
A beam is supported at its ends by supports which are 12 metres
apart. Since the load is concentrated at its centre, there is a
deflection of 3 cm at the centre and the deflected beam is in the
shape of a parabola. How far from the centre is the deflection 1 cm?
Y
12m
AO epee
x’ xX
D
y’
Beam is always facing upwards with the axis vertical
Since, the axis is positive y-axis,
its equation is
x? = day
Example, 19
Example 19 teachoo.com
Arod AB of length 15 cm rests in between two coordinate axes in
such a way that the end point A lies on x-axis and end point B lies
on y-axis. A point P(x, y) is taken on the rod in such a way that AP =
6 cm. Show that the locus of P is an ellipse.
Y
Given AB=15 cm B
&AP=6cem 2 em
QP P(x, y)
Now, “ | 6cm x
0 R A
PB=AB-—AP ,
Y
PB=15-6
PB=9cm
Drawing PQ L BO and PR L OA
Miscellaneous
8 questionsMisc 1
Misc 1 teachoo.com
If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the
focus. Y
_cm _ a
Let equation of parabola be
y?= 4ax 20cm
rs) xX
We need to find focus
B
To find focus, we first find coordinates of point A,
And, then put it’s coordinates in equation of parabola
Misc 2
Misc 2 teachoo.com
Anarch is in the form of a parabola with its axis vertical. The arch is
10 m high and 5 m wide at the base. How wide is it 2m from the
vertex of the parabola?
Y
18]
x’ xX
Arch is downwards,
Since, the axis of parabola is negative y-axis, 10m
its equation is
a Cc -=
x? = —day A B
sm
y’
First, we find coordinates of point B
Given,
Width of Parabola = AB = 5m
Misc 3
. teachoo.com
Misc 3 | ] | ]
nil ill tee
fll
The cable of a uniformly loaded suspension bridge hangs in the form
of a parabola. The roadway which is horizontal and 100 m long is
supported by vertical wires attached to the cable, the longest wire
being 30 m and the shortest being 6 m. Find the length of a
supporting wire attached to the roadway 18 m from the middle.
Y
100 m
Ae-------------£ ---------------=B
A “om
xX’ xX
y
Cable is always facing upwards with the axis vertical
Misc 4
Misc 4 teachoo.com
An arch is in the form of semi ellipse. It is 8m wide and 2 m high at
the centre. Find the height of the arch at a point 1.5 m from one end.
Y
The arch is inform of semi-ellipse C
It is 8m wide & 2m high
. ae Xx’ X|
Let AB = width of semi-ellipse = 8m Al . 3 7B
7
& CO = Height of semi-ellipse = 2m SA 7
Y’
We know that
Length of Major axis of ellipse = 2a | Length of semi-minor axis = b
AB =2a CcO=b
8=2a 2=b
2a=8 be2
as224
2
Misc, 5
Misc S teachoo.com
A rod of length 12 cm moves with its ends always touching the
coordinate axes. Determine the equation of the locus of a point P
on the rod, which is 3 cm from the end in contact with the x-axis.
Y
Given AB=12 cm
B
& AP=3cm cm
aP P(x, y)
Now,
3cm
ape 1 D x
PB = AB-AP XA R A
PB=12-3 Y’
PB=9cem
Drawing PQ | BO and PR L OA
Misc 6
Misc 6 teackoo.com
Find the area of the triangle formed by the lines joining the vertex
of the parabola x? = 12y to the ends of its latus rectum.
Y x? = 12y
The given equation of parabola is
x? = 12y (1)
The above equation is of the form x x
x? = day (2) 0
y
Comparing (1) & (2)
12y = day
12y
"sa
ay
3=a
a=3
Misc 7
Misc 7 (Method 1) teachoo.com
A man running a racecourse notes that the sum of the distances
from the two flag posts from him is always 10 m and the distance
between the flag posts is 8 m. find the equation of the posts traced
by the man.
Y
Let man be standing on point P(x, y)} (15 P(x, y)
There are two flag postsS & S ” ata,/ \ x
Given
&SS’=8m y’
Let S & S’ be on x-axis
such that Origin (O) be the mid-point of S’S
So, OS=0S’=4m
Misc 8
Misc 8 teachoo.com
An equilateral triangle is inscribed in the parabola y? = 4ax, where
one vertex is at the vertex of the parabola. Find the length of the
side of the triangle.
Y
Buy? = 4ax
Let length of equilateral triangle be s
Ss
Hence, OA = OB = AB=s 2
xX’ xX
oO Cc
5
Here, OC | AB 2
A
So, Z OCA = Z OCB = 90° y’
AndAC=BC (Since parabola is symmetric about its axis)
So, AC = BC =
s
AC=BC= 3
Why Learn This With Teachoo?
Conic Sections studies the curves obtained when a plane intersects a double cone: circle, parabola, ellipse and hyperbola. Students learn their geometric definitions, standard equations and key elements such as focus, directrix, vertex, centre, axis, eccentricity and latus rectum. Teachoo provides NCERT solutions, examples, miscellaneous questions and concept-wise practice for Class 11 Conic Sections, including applied problems involving arches, mirrors, beams, loci and paths.
How are conic sections defined?
A conic can be described as the locus of a point whose distance from a fixed point, called the focus, bears a constant ratio e to its perpendicular distance from a fixed line, called the directrix. The constant e is eccentricity:
-
e = 0 for a circle in the centred interpretation;
-
e = 1 for a parabola;
-
0 < e < 1 for an ellipse;
-
e > 1 for a hyperbola.
The NCERT chapter develops standard forms aligned with coordinate axes. Students should sketch each curve and label its elements before applying formulas.
Circle
A circle is the locus of a point at a constant distance r from a fixed centre (h, k). Its standard equation is
(x − h)² + (y − k)² = r².
The special form x² + y² = r² has centre at the origin. Expanding or completing squares helps identify a circle from a general equation and obtain its centre and radius.
Parabola
A parabola is the locus of a point equidistant from a fixed focus and a fixed directrix. Standard equations include y² = 4ax, y² = −4ax, x² = 4ay and x² = −4ay. For y² = 4ax, the vertex is (0, 0), focus is (a, 0), directrix is x = −a and latus rectum length is 4a.
Orientation depends on which variable is squared and the sign of the unsquared term. Parabolic models appear in arches, reflectors, headlights and satellite dishes.
Ellipse
An ellipse is the locus of a point for which the sum of distances from two fixed foci is constant. For x²/a² + y²/b² = 1 with a > b > 0, the major axis lies on the x-axis, c² = a² − b² and eccentricity e = c/a. The foci are (±c, 0), vertices are (±a, 0) and the latus rectum length is 2b²/a.
If the larger denominator lies under y², the major axis is vertical. Students must identify a and b from size, not automatically from variable names.
Hyperbola
A hyperbola is the locus of a point for which the absolute difference of distances from two fixed foci is constant. For x²/a² − y²/b² = 1, c² = a² + b², e = c/a, foci are (±c, 0) and vertices are (±a, 0). For y²/a² − x²/b² = 1, the transverse axis is vertical. The sign of the positive squared term indicates the opening direction.
Topics covered on Teachoo
-
Exercises 10.1 to 10.4, examples and miscellaneous questions;
-
circle equations and parameters;
-
basic parabola questions;
-
parabola applications involving arches and mirrors;
-
triangle and beam problems in a parabola;
-
ellipse definition and standard results;
-
ellipse as a locus and path-tracing problems;
-
hyperbola and its standard equations;
-
identification of focus, directrix, vertex, axes and latus rectum.
Learning outcomes
Students should be able to classify a conic from its equation, sketch its orientation and identify its parameters. They should form a standard equation from geometric data, find foci, vertices, eccentricity, directrices and latus rectum, and solve application or locus problems. They should complete squares to recognise translated circles.
Why is this chapter important?
Conics unite algebra, geometry and real-world modelling. Planetary paths, reflectors, optical devices, arches and navigation use conic properties. The chapter has strong JEE relevance and prepares students for tangents, normals and coordinate geometry in higher classes.
How Teachoo helps you prepare
Teachoo groups questions by conic and by application, so students can build one curve at a time. Make a four-column summary for equation, orientation, focus and directrix. For every question, sketch first, locate the axis and only then choose a formula.
Use serial-order NCERT solutions to complete exercises and examples, and concept-wise lessons for application questions. In a locus derivation, begin from the geometric distance condition, use the distance formula, simplify carefully and interpret the resulting equation.
School-exam, JEE and competency preparation
School exams test standard equations and parameter identification. JEE questions often combine conics with lines, parameters or loci. Do not memorise focus formulas without associating them with a labelled sketch.
Competency questions may supply the dimensions of an arch or the location of a light source. Choose the origin and axes to exploit symmetry, translate the dimensions into coordinates and substitute them into the appropriate standard equation. State units in the final application answer.
Quick revision checklist
Sketch all four conics; label vertices, foci, directrices and axes; identify a conic from five equations; form one equation of each type; complete squares for a circle; and solve one parabola and one ellipse application.
Common mistakes to avoid
Do not decide ellipse axes from the variable alone—look for the larger denominator. For a hyperbola, use c² = a² + b², not a² − b². In a parabola, the focus is a units from the vertex while the latus rectum length is 4a. Keep a² and a distinct. A rough graph should reflect the equation’s opening direction.
Deeper reasoning and concept connections
In Conic Sections, fluency means more than repeating a procedure. Students should be able to recognise the underlying structure when the numbers, diagram, wording or orientation changes. A useful routine is: identify the mathematical objects, list the known and unknown quantities, state the governing property, carry out the steps and verify that every condition has been used.
Look for connections within the chapter as well. A definition usually leads to a representation; the representation reveals a pattern; and the pattern supports a rule or calculation. Explaining this chain improves retention and helps with case-based questions. It also prevents the common mistake of selecting a formula simply because its symbols resemble the numbers in the question.
How to solve unfamiliar and competency-based questions
Use a five-step response: interpret, represent, select, solve and verify. Interpret the wording; represent the information; select a definition, property or formula; solve without skipping the logical step; and verify through substitution, estimation, measurement or an alternative representation. This routine works for direct exercises as well as case-based questions.
If information appears unnecessary, ask whether it establishes a hidden condition. If information is missing, state what cannot be determined instead of inventing a value. In written answers, name the rule being used. Clear reasoning helps a teacher award method marks and also makes the page easier for a student—or an AI answer system—to retrieve for the precise doubt being asked.
What complete mastery looks like
For Conic Sections, a student should be able to define the central ideas in simple language, recognise them in different representations, solve routine questions accurately and explain the method used. They should also be able to correct a flawed solution, create an example satisfying given conditions and combine two ideas from the chapter in one problem. A reliable mastery test is to solve one direct question, one application question and one reasoning question without looking at notes, then explain all three aloud or in writing.
Keep a compact error log with four labels: concept, interpretation, calculation and presentation. Reattempt each error after a gap instead of rereading the answer immediately. Improvement comes from correcting the decision that caused the mistake, not from repeating questions whose method is already known.
Additional frequently asked questions
What should a student know before starting Conic Sections?
Revise the definitions, number operations, diagrams or notation used at the beginning of the chapter. The prerequisite list should be short: if an earlier skill blocks progress, repair that skill with two or three focused questions and return to the chapter.
How can a student check an answer in Conic Sections?
Use an independent check whenever possible: substitute the result, reverse the operation, estimate its size, compare it with the figure, test a simpler case or solve using another representation. A check should examine the mathematical condition, not merely repeat the same arithmetic.
How many questions are enough for strong preparation?
There is no fixed number. Stop counting questions and track coverage: every concept, every standard method, at least one mixed problem, one competency-based problem and every previously incorrect type should be solved independently. Ten varied, analysed questions are more valuable than fifty copied solutions.
How should Teachoo solutions be used without becoming dependent on them?
Attempt the question first and mark the exact step where progress stops. Read only enough of the solution to repair that step, close it and restart the question. Finally, solve a similar problem without help. This turns a solution into feedback rather than a substitute for thinking.
Frequently asked questions
How do I identify a conic from its standard equation?
One squared variable suggests a parabola; two same-sign squared terms suggest an ellipse or circle; opposite-sign squared terms suggest a hyperbola.
What is eccentricity?
It is the constant ratio of distance from a focus to perpendicular distance from a directrix in the focus-directrix definition.
How do I know the orientation of a parabola?
The squared variable identifies the axis perpendicular to it, while the sign of the unsquared term determines the positive or negative opening direction.
What is the main difference between an ellipse and a hyperbola?
An ellipse has a constant sum of focal distances; a hyperbola has a constant absolute difference.
Does Teachoo include applied conic questions?
Yes. Teachoo includes parabola problems involving arches, mirrors, triangles and beams, plus ellipse locus and path-tracing problems.
Treat each standard equation as a geometric description. A labelled sketch makes the formulas meaningful and sharply reduces sign and orientation errors.